Explanation
We must find the product of complex numbers and write it in rectangular form i.e. like a+ib where a and b are real numbers. In order to do tthis we first need to write the values of the sine and cosine of 120° and 210°. For this purpose we can use a table displaying the values of the trigonometric functions for several angles. However, these tables usually display the values for angles between 0° and 90° like the following one:
However this table is still useful. 120° is an angle in the second quadrant which means that it has the same sine as an angle of the first quadrant that meets x=180°-120° and its cosine is the same but multiplied by -1. We have x=180°-120°=60° which means that:
[tex]\begin{gathered} \sin120^{\circ}=\sin60^{\circ}=\frac{\sqrt{3}}{2} \\ \cos120^{\circ}=-\cos60^{\circ}=-\frac{1}{2} \end{gathered}[/tex]
We can do something similar for 210°. It's in the third quadrant which means that both its sine and cosine are thoes of x=210°-180°=30° but multiplied by -1. Then we get:
[tex]\begin{gathered} \sin210^{\circ}=-\sin30^{\circ}=-\frac{1}{2} \\ \cos210^{\circ}=-\cos30^{\circ}=-\frac{\sqrt{3}}{2} \end{gathered}[/tex]
Now that we found the sine and cosine of both angles we can rewrite the product:
[tex]\begin{gathered} \lbrack4(\cos210^{\circ}+i\sin210^{\circ})\rbrack\cdot\lbrack5(\cos(120^{\operatorname{\circ}})+\imaginaryI\sin(120^{\operatorname{\circ}}))\rbrack= \\ =\lbrack4(-\frac{\sqrt{3}}{2}-i\frac{1}{2})\rbrack\cdot\lbrack5(-\frac{1}{2}+\mathrm{i}\frac{\sqrt{3}}{2})\rbrack=\lbrack-2\sqrt{3}-2i\rbrack\cdot\lbrack-\frac{5}{2}+\mathrm{i}\frac{5\sqrt{3}}{2}\rbrack \end{gathered}[/tex]
Then we get:
[tex]\begin{gathered} \lbrack-2\sqrt{3}-2i\rbrack\cdot\lbrack-\frac{5}{2}+\mathrm{i}\frac{5\sqrt{3}}{2}\rbrack=2\sqrt{3}\cdot\frac{5}{2}-i2\sqrt{3}\cdot\frac{5\sqrt{3}}{2}+i2\cdot\frac{5}{2}-i\cdot i\cdot2\cdot\frac{5\sqrt{3}}{2} \\ =5\sqrt{3}-i15+i5+5\sqrt{3}=10\sqrt{3}-10i \end{gathered}[/tex]Answer
Then the answer is:
[tex]10\sqrt{3}-i10[/tex]
